ON CONVEX REPRESENTATIONS OF MAXIMAL MONOTONE OPERATORS IN NON-REFLEXIVE BANACH SPACES

Transcription

1 INSTITUTO NACIONAL DE MATEMÁTICA PURA E APLICADA ON CONVEX REPRESENTATIONS OF MAXIMAL MONOTONE OPERATORS IN NON-REFLEXIVE BANACH SPACES REPRESENTAÇÕES CONVEXAS DE OPERADORES MONÓTONOS MAXIMAIS EM ESPAÇOS DE BANACH NÃO REFLEXIVOS Author: Maicon Marques Alves Adviser: Prof. Dr. Benar Fux Svaiter February 2009

2 Livros Grátis Milhares de livros grátis para download.

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4 ii To my wife Daiane

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6 Abstract The main focus of this thesis is to study some generalizations of a theorem of Burachik and Svaiter to non-reflexive Banach spaces, and its theoretical implications in the theory of maximal monotone operators, with special emphasis for maximal monotone operators of type (NI). Keywords: Maximal monotone operators, convex functions, convex representations, non-reflexive Banach spaces, Fitzpatrick functions, Fitzpatrick family, operators of type (NI), extension to the bidual, duality mapping, Gossez generalized duality mapping, surjectivity. iv

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8 Agradecimentos Muitas pessoas contribuíram para que fosse possível chegar até aqui. Começo então agradecendo à minha esposa Daiane, que dedicou muito amor e companheirismo ao longo desses anos, e minha mãe Aldalete, que sempre priorizou os meus estudos. A toda minha família e ao meu irmão querido, também agradeço por tudo. Ao meu orientador Professor Benar Fux Svaiter, meus sinceros agradecimentos por sua serenidade, paciência e competência demonstrados nesse período. Esta tese é um resultado de longas horas de trabalho conjunto que realizamos. Aos Professores Antonio Leitão e Mario Rocha Retamoso agradeço pelos incentivos no mestrado e na graduação, pela acolhida sempre atensiosa quando volto as origens, pelas oportunidades que me proporcionaram e pelos ensinamentos transmitidos. Agradeço também ao Professor Alfredo Iusem por permitir o meu ingresso no Grupo de Otimização do IMPA e pelas conversas sempre esclarecedoras. Agradeço aos Professores Alfredo Iusem, Fernanda Raupp, Paulo Klinger, Rolando Gárciga Otero e Susana Scheimberg por parciciparem da comissão examinadora. A todos os amigos e colegas fica o meu muito obrigado por tudo. Em especial agradeço à Adriano, André, Cleverson, Fábio Júlio, Ives, Jean, Jefferson, Vanderlei, Sérgio e Yunier pela amizade, pelas conversas matemáticas e pelos muitos momentos de descontração proporcionados. Finalmente gostaria de agradecer à todos os professores e funcionários do IMPA. Em particular agradeço a Andrea Pereira e Rogerio de Souza Silva pela forma simpática e competente que sempre me receberam. Ao CNPq agradeço pelo apoio financeiro. vi

9 Contents Introduction 1 1 Basic results and notation Fitzpatrick functions The starting point Maximality in non-reflexive Banach spaces Maximality of subdifferentials Maximality and Brøndsted-Rockafellar property Preliminary results Maximality of representable monotone operators Brøndsted-Rockafellar type theorems for representable and of type (NI) operators A new characterization and a sum theorem for operators of type (NI) On the relation between surjectivity of perturbations and operators of type (NI) Preliminary results Main results On the uniqueness of the extension to the bidual Convexity and maximal monotonicity Basic results and some notation Extension theorems A Basics of convex analysis 49 vii

10 Introduction Maximal monotone operators appear in several branches of applied mathematics such as optimization, partial differential equations, and variational analysis. These operators were object of intense research between 1960 and 1980, when Brezis, Browder, Minty and Rockafellar established the fundamental results about them. The theory of convex representations of maximal monotone operators emerged at the end of the 1980s with the work [20] of S. Fitzpatrick 1. It took some years until the Fitzpatrick s results were rediscovered by Martínez-Legaz and Théra [35] and Burachik and Svaiter [17]. Since then, this subject has called the attention of several researchers and has been object of intense research in the field of convex analysis, monotone operator theory and optimization [2, 3, 4, 11, 7, 9, 8, 10, 18, 33, 34, 31, 30, 29, 32, 36, 43]. We refer the reader to [17] for exploiting the relationship between convex representations and enlargements of maximal monotone operators. The main point on Fitzpatrick s result is that it allows the use of convex analysis in the study of maximal monotone operators. The proof of many basic results can be simplified and also new results obtained, mostly in non-reflexive Banach spaces, with this new tool. In this thesis we will present some of the new results on maximal monotone operators, as well as a simple proof of Rockafellar theorem on the maximality of the subdifferential. This thesis is build around the generalization of a Burachik-Svaiter theorem about maximal monotonicity on reflexive Banach spaces [18]. We generalized this result to non-reflexive Banach spaces. Together with some techniques used for this generalization it allowed us to obtain some other results, which are also presented here. A classical theorem of Rockafellar states that the subdifferential is maximal monotone. Unfortunately, most of the techniques Rockafellar used for proving this fact could not be adapted to our aims. For these reasons, a new and simpler proof of the maximal monotonicity of the subdifferential was the first result we obtained

11 Presentation of chapters Chapter 1: In this chapter we present the results of Fitzpatrick and Burachik-Svaiter on convex representations of maximal monotone operators. Fitzpatrick s results are summarized in Theorem In Section 1.2, we called The starting point, Burachik-Svaiter s results in this direction of research are presented. Chapter 2: This chapter is the first step toward the study of maximal monotonicity in non-reflexive Banach spaces. We start, in Section 2.1, by presenting a new proof of maximality of the subdifferential of a convex function. This proof is simpler than Rockafellar s classical proof, and makes use of classical results from subdifferential calculus as Brøndsted-Rockafellar s Theorem and Fenchel- Rockafellar duality Theorem. We also observe that the proof can be simplified in reflexive spaces, and that it can be seen as a particular case of a more general maximality result presented in Theorem Section 2.2 is devoted to the study of monotone operators representable by convex functions satisfying condition (1.13). Theorem 2.2.5, states that condition (1.13) is a sufficient condition for maximal monotonicity in non-reflexive spaces, generalizing Theorem for this non-reflexive setting. In Theorem and Theorem 2.2.8, we shall prove that such operators satisfies a Brøndsted- Rockafellar type property. In this context, we futher prove, in Theorem , that maximal monotone operators of type (NI) also satisfies this Brøndsted-Rockafellar type property. In Theorem and Lemma 2.3.2, of Section 2.3, we prove some additional properties of operators of type (NI) and a sum theorem for this class of operators, respectively. The results of this chapter were published in [28, 29, 31]. Chapter 3: This chapter is about surjectivity of perturbation of maximal monotone operators in non-reflexive Banach spaces. The results presented are from [30]. In a reflexive Banach space surjectivity of a monotone operator plus the duality mapping is equivalent to maximal monotonicity. This is a classical result of Rockafellar [38]. In [21] Gossez introduced an enlarged version of the duality mapping and proved similar Rockafellar s surjectivity results for the class of maximal monotone operators of type (D), introduced by himself. The class of maximal monotone operators of type (NI), introduced by S. Simons in [39], encompasses the Gossez type (D) operators. 2

12 We shall make use of the analytical tools developted in Chapter 2 in order to obtain surjectivity results of perturbations of operators of type (NI) by using the Gossez enlarged duality mapping. The main results of the Chapter are present in Theorem Chapter 4: Any maximal monotone operator T : X X is also a monotone operator T : X X and admits one (or more) maximal monotone extension in X X that (in general) will not be unique. This chapter approaches the problem of under which conditions a maximal monotone operator T : X X has a unique extension to the bidual. The Gossez type (D) maximal monotone operators have a unique maximal monotone extension to the bidual [21, 22, 23, 24]. We will prove that maximal monotone operators of type (NI) admit a unique extension to the bidual and that, for non-linear operators, the condition (NI) is equivalent to the unicity of maximal monotone extension to the bidual. For proving this equivalence we will show that if T X X is maximal monotone and convex then T is an affine subspace. The results of this chapter are from the paper [31]. Notations B X [0,M] closed ball of X with radius M cl f largest l.s.c. function majorized by f conv f largest convex function majorized by f cl conv f largest l.s.c. convex function majorized by f dom(f) effective domain of f δ A indicator function of A ϕ T Fitzpatrick function of T F T Fitzpatrick family of T J duality mapping J = J ε enlarged duality mapping J ε = ε 2 2 Λh Λh(x,x ) = h (x,x) π duality product in X X π duality product in X X R extended real system { } R { } R X set of functions of X into R R(T) range of T R R : X X X X, R(x,x ) = (x,x ) S T S-function of T T 1 inverse of T 3

13 Chapter 1 Basic results and notation Let X be a real Banach space with topological dual X. For x X and x X we will use the notation x,x = x (x). A point to set operator T : X X is a relation on X X : T X X and x T(x) means (x,x ) T. An operator T : X X is monotone if x y,x y 0, (x,x ), (y,y ) T and it is maximal monotone if it is monotone and maximal (with respect to the inclusion) in the family of monotone operators of X into X. 1.1 Fitzpatrick functions Brezis and Haraux defined in [12], for a maximal monotone operator T : X X, the function β T R X X, β T (x,x ) = sup x y,y x. (1.1) (y,y ) T Note that if (x,x ) T then the above inner product is always nonpositive, being equal to zero for (y,y ) = (x,x ). So, β T = 0 in T. If (x,x ) / T, since T is maximal monotone we conclude that x y,x y < 0 for some (y,y ) T and so β T (x,x ) > 0. Therefore, for any (x,x ) X X β T (x,x ) 0, β T (x,x ) = 0 (x,x ) T. The function β T provides a representation of the maximal monotone operator, but it lacks properties to be explored. 4

14 Fitzpatrick defined [20], for a maximal monotone operator T, the function ϕ T (x,x ) = sup x y,y x + x,x (1.2) (y,y ) T = sup (y,y ) T y,x + x,y y,y. (1.3) Using (1.2) and the previous observations we conclude that ϕ T (x,x ) x,x, ϕ T (x,x ) = x,x (x,x ) T. (1.4) Note also from (1.3) that ϕ T, being a sup of linear functions on (x,x ), is convex and lowersemicontinuous. The above equation generalizes, in some sense, Fenchel-Young inequality (2.1). Fitzpatrick also defined a family of convex functions associated with each maximal monotone operator T, h is convex and lower semicontinuous X X F T = h R h(x,x ) x,x, (x,x ) X X (x,x ) T h(x,x ) = x,x. (1.5) and proved the next result: Theorem ([20, Theorem 3.10]). Let T : X X be maximal monotone. Then for any h F T (x,x ) T h(x,x ) = x,x (1.6) and ϕ T is the smallest element of the family F T. Proof. Inclusion of ϕ T in F T has already been proved in (1.4), and in the preceding discussion. To prove that ϕ T is minimal in F T, take h F T and (y,y ) T. For any (x,x ) and p,q 0, p + q = 1, we have px + qy,px + qy h(px + qy,px + qy ) ph(x,x ) + qh(y,y ) (1.7) where the first inequality follows for the fact that h majorizes the duality product and the second one from the convexity of h. Since (y,y ) T, h(y,y ) = y,y, which combined with (1.7) yields p x,x + q[ x,y + y,x y,y ] h(x,x ). Now, taking limit as p 0 (q 1) in the above inequality and supremum over (y,y ) T in the resulting inequality we conclude that ϕ T h. Therefore, ϕ T is minimal in F T and (1.6) follows from this minimality. 5

15 It is worth to note that by (1.6), each Fitzpatrick function fully characterizes the operator T which defines the family F T. This is exactly what we mean by a convex representation of a maximal monotone operator T. For instance, if f is a proper, lower semicontinuous convex function on X, then h s F f, where h s (x,x ) = f(x) + f (x ). 1.2 The starting point In this section we will discuss previous results of Burachik and Svaiter, which are the starting point of this thesis. Fitzpatrick function combines duality product and conjugation. Before showing that, we will establish some notation and conventions. Recall that the conjugate of f R X is f R X, f (x ) = sup x,x f(x). (1.8) x X From now on we will denote the duality product by π, π R X X, π(x,x ) = x,x. (1.9) We will also identify X with its image under the canonical injection into X. With these conventions, we have ϕ T (x,x ) = (π + δ T ) (x,x) where δ T R X X is the indicator function of T: { 0, on T,, otherwise In [17] Burachik and Svaiter observed that the family F T is closed under the supremum operation and defined its largest element, the so called S-function: S T R X X, S T = sup h F T h, (1.10) Hence, S T F T. The next theorem gives an explicit expression of S T and use this function (together with ϕ T ) to provide an alternative characterization of F T (to be used in Theorem 4.3.1). Theorem ([17, Corollary 4.1, Remark 5.4]). Let ϕ T and S T be the Fitzpatrick and S-function associated to T, respectively, as defined in (1.2) and (1.10). Then, 6

16 1. S T = cl conv(π + δ T ), 2. for any convex lower semicontinuous function h R X X, h F T ϕ T h S T, 3. for any (x,x ) X X, ϕ T (x,x ) = (S T ) (x,x), 4. if X is reflexive, then for any (x,x ) X X S T (x,x ) = (ϕ T ) (x,x). If we define, Λ : R X X R X X, Λh(x,x ) := h (x,x), (1.11) according to the above theorem, ΛS T = ϕ T F T. So, it is natural to ask whether Λ maps F T into itself. Burachik and Svaiter also proved that this happens in fact: Theorem ([17, Theorem 5.3]). Suppose that T is maximal monotone. Then Λh F T, h F T, that is, if h F T, and g : X X R, g(x,x ) = h (x,x), then g F T. In a reflexive Banach space Λϕ T = S T. It is interesting to note that Λ is an order-reversing mapping of F T into itself. This fact suggests that this mapping may have fixed points in F T. Svaiter proved [43] that if T is maximal monotone, then Λ always has a fixed point in F T. Note that, by Theorem 1.2.2, if h F T then: h(x,x ) x,x, (x,x ) X X, h (x,x) x,x, (x,x ) X X. (1.12) In [18] Burachik and Svaiter proved that the converse of this implication holds in a reflexive Banach space: Theorem ([18, Theorem 3.1]). Let X be a reflexive Banach X X space. If h R is proper, convex, l.s.c. and then h (x,x ) x,x, (x,x ) X X h (x,x) x,x, (x,x ) X X T := {(x,x ) X X h (x,x ) = x,x } is maximal monotone and h, Λh F T. 7

17 The proof of Theorem was based on Rockafellar s surjectivity theorem (see Theorem 3.0.3). Unfortunately, this technique cannot be used in a non-reflexive setting. Motivated by this problem, in [29] we propose to replace (1.12) by the novel condition: h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X. (1.13) The above condition generalizes (1.12) and allows one to prove Theorem in non-reflexive spaces, by replacing condition (1.12) by (1.13) and by using the Fenchel-Rockafellar duality Theorem A.0.4. It still remains as an open question to prove Theorem in nonreflexive Banach spaces with the original Burachik-Svaiter s condition (1.12). In the last years, a great effort has been spent in trying to answer affirmatively such open question. Theorem would give a affirmative answer to the celebrate Rockafellar s conjecture on the sum of maximal monotone operators in non-reflexive Banach spaces (see Lemma 2.3.2). The main focus of this thesis is to study some generalizations of Theorem to non-reflexive Banach spaces, and its theoretical implications in the theory of maximal monotone operators, with special emphasis for maximal monotone operators of type (NI). The results presented in this thesis are contained in the works [28, 29, 30, 31, 32], all of them in collaboration with B. F. Svaiter. 8

18 Chapter 2 Maximality in non-reflexive Banach spaces This chapter is concerned with the study of maximal monotonicity in non-reflexive Banach spaces. In Section 2.1, we give a new proof of the maximality of the subdifferential of a convex function. Section 2.2 is devoted to the study of monotone operators representable by convex functions satisfying condition (1.13). In Section 2.3 we prove some additional properties of operators of type (NI) and a sum theorem for this class of operators. The results of this chapter were published in [28, 29, 31]. 2.1 Maximality of subdifferentials Recall that the subdifferential of f is the operator f : X X, f(x) = {x X f(y) f(x) + y x,x, y X}. Using the above definition, it is easy to check that if f is proper, convex and lower semicontinuous (l.s.c. for short), then f is monotone and f(x) + f (x ) x,x, f(x) + f (x ) = x,x x f(x). (2.1) First proved by Rockafellar in [37], the maximal monotonicity of the subdifferential of a convex function is still object of study and several authors attempt to give simpler proofs to this fact (see [28] and references therein). Rockafellar s original proof is based on very important tools introduced by himself in [37]. In particular, he has proved a result of weak density for the graph of f in the graph of f that has been widely used in different situations in convex 9

19 analysis (see Theorem 6.1 of [17] for an application) and later on was introduced by Gossez in the context of maximal monotone operators of type (D) [23, 24, 21, 22]. In this section we present a short proof for the maximality of subdifferentials which makes use of classical results from subdifferential calculus like Brøndsted-Rockafellar s Theorem (Theorem A.0.3) and Fenchel-Rockafellar duality Theorem (Theorem A.0.4). We also observe that our proof can be still simplified in reflexive spaces, in particular in finite dimensional spaces, and it can be seen as a particular case of a more general maximality result presented in Theorem Theorem Let X be a Banach space. If f R X is proper, convex and l.s.c., then f : X X is maximal monotone. Proof. (Marques Alves-Svaiter) Monotonicity of f is easy to check. Suppose that (x 0,x 0) X X is such that Define x x 0,x x 0 0, x f(x). f 0 R X, f 0 (x) := f(x + x 0 ) x,x 0. (2.2) Applying Theorem A.0.4 to f 0 and g(x) := 1 2 x 2 we conclude that there exists x X such that inf f 0(x) + 1 x X 2 x 2 + f0(x ) x 2 = 0. (2.3) In particular, there exists a (minimizing) sequence {x n } such that 1 n 2 f 0 (x n ) x n 2 + f 0(x ) x 2 x n,x x n x ( x n x ) 2 0, (2.4) where the second inequality follows from Fenchel-Young inequality. Using the above equation we obtain f 0 (x n ) + f 0(x ) x n,x 1/n 2. Hence, x 1/n 2f 0 (x n ) and by Theorem A.0.3 it follows that there exist sequences {z n } in X and {z n} in X such that z n f 0 (z n ), z n x 1/n and z n x n 1/n. (2.5) Using the initial assumption, we also obtain z n,z n 0. (2.6) 10

20 Using (2.4) we get x n x, x n,x x 2, as n, (2.7) which, combined with (2.5) and (2.6) yields x = 0. Therefore, x n 0. As f 0 is l.s.c., x = 0 minimizes f 0 (x) x 2 and, using (2.3) we have f 0 (0) + f0(0) = 0. Therefore 0 f 0 (0), which is equivalent to x 0 f(x 0 ). Notice that in a reflexive Banach space X (in particular, in finite dimensional vector spaces) the proof of Theorem can be futher simplified by taking a minimum on (2.3). This leads to the existence of z X such that f 0 (z) + f 0(x ) = z,x, 1 2 z x 2 + z,x = 0 and so 0 f 0 (0), which finishes the proof. It should also be noted that a similar (and simple) proof for Theorem can be obtained directly from Theorem by using h s (x,x ) := f(x) + f (x ) as a convex representation for the monotone operator f. 2.2 Maximality and Brøndsted-Rockafellar property In this section we are interested in the study of maximality of monotone operators representable by convex functions satisfying condition (1.13). A remarkable result is Theorem 2.2.5, in which we prove that condition (1.13) is a sufficient condition for maximal monotonicity in non-reflexive spaces, generalizing Theorem for this non-reflexive setting. After proving Theorem we will study Brøndsted-Rockafellar property for maximal monotone operators in non-reflexive Banach spaces. Burachik, Iusem and Svaiter [15] defined the ε-enlargement of T for ε 0, as T ε : X X T ε (x) = {x X x y,x y ε (y,y ) T }. (2.8) It is trivial to verify that T T ε. The ε-enlargement is a generalization of the ε-subdifferential of a convex function and has both theoretical and practical uses [41, 42, 19, 25, 26, 27]. An important question concerning the study of ε-enlargements of a maximal 11

21 monotone operator T is whether an element in the graph of T ε can be approximated by an element in the graph of T. A maximal monotone operator T : X X has the Brøndsted- Rockafellar property if, for any ε > 0, x T ε (x) λ > 0, ( x, x ) T, x x λ, x x ε/λ. It does make sense to ask if every maximal monotone operator has Brøndsted-Rockafellar property. This question has been successfully solved for the extension ε f, of f, by Brønsted and Rockafellar, as is showed in Theorem A.0.3. In the case of a general maximal monotone operator, the answer is affirmative in reflexive Banach spaces [44, 16] but is negative in the non-reflexive case [40]. The operator T satisfies the strict Brøndsted-Rockafellar property [29] if x T ε (x), η > ε λ > 0, ( x, x ) T, x x < λ, x x < η/λ. (2.9) In Theorem and Theorem we will prove that maximal monotone operators representable by convex functions satisfying (1.13) and maximal monotone operators of type (NI) satisfies the strict Brøndsted-Rockafellar property. The results of this section were published in [29, 31] Preliminary results The main result of this subsection is Theorem We start by proving some technical results. X X Theorem (Marques Alves-Svaiter [29]) Let h R be a convex and l.s.c. function. If h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, then for any ε > 0 there exists ( x, x ) X X such that h( x, x )+ 1 2 x x 2 < ε, x 2 h(0, 0), x 2 h(0, 0), where the two last inequalities are strict in the case h(0, 0) > 0. Proof. If h(0, 0) < ε then ( x, x ) = (0, 0) has the desired properties. The non-trivial case is ε h(0, 0), (2.10) 12

22 which we consider now. Using the assumptions on h, we conclude that for any (x,x ) X X, h(x,x ) x x 2 x,x x x 2 x x x x 2 Analogously, for all (z,z ) X X, = 1 2 ( x x ) 2 0. (2.11) h(z,z ) z z 2 z,z z z 2 z z z z 2 = 1 2 ( z z ) 2 0. (2.12) Now using Theorem A.0.4 for the Banach space X X and f,g R X X, f(x,x ) := h(x,x ), g(x,x ) := 1 2 x x 2 we conclude that there exists (ẑ,ẑ ) X X such that inf h(x,x ) + 1 (x,x ) 2 x x 2 = h (ẑ,ẑ ) 1 2 ẑ ẑ 2. As the right hand side of the above equation is non positive and the left hand side is non negative, these two terms are zero. Therefore, inf h(x,x ) + 1 (x,x ) X X 2 x x 2 = 0, (2.13) and h (ẑ,ẑ ) ẑ ẑ 2 = 0. (2.14) For (z,z ) = (ẑ,ẑ ), all inequalities on (2.12) must hold as equalities. Therefore, ẑ 2 = ẑ 2 = h(ẑ,ẑ ) h(0, 0), (2.15) where the last inequality follows from the definition of conjugate. 13

23 Using (2.13) we conclude that for any η > 0, there exists (x η,x η ) X X such that h(x η,x η ) x η x η 2 < η. (2.16) If h(0, 0) =, then, taking η = ε and ( x, x ) = (x η,x η ) we conclude that the theorem holds. Now, we discuss the case h(0, 0) <. In this case, using (2.15) we have Note that from (2.10) we are considering ẑ = ẑ h(0, 0). (2.17) ε h(0, 0) <. (2.18) Combining (2.14) with (2.16) and using Fenchel-Young inequality (A.2) we obtain η > h(x η,x η ) x η x η 2 + h (ẑ,ẑ ) ẑ ẑ 2 x η,ẑ + x η,ẑ x η x η ẑ ẑ x η 2 x η ẑ ẑ x η 2 x η ẑ ẑ 2 = 1 2 ( x η ẑ ) ( x η ẑ ) 2. As the two terms in the last inequality are non negative, x η < ẑ + 2η, x η < ẑ + 2η. Therefore, using (2.17) we obtain x η < h(0, 0) + 2η, x η < h(0, 0) + 2η. For finishing the proof, take in (2.16) and let τ = 0 < η < ε 2 2h(0, 0) (2.19) h(0, 0) h(0, 0) + 2η x = τ x η, x = τ x η. (2.20) Then, x < h(0, 0), x < h(0, 0). 14

24 Now, using the convexity of h and of the square of the norms and (2.16), we have h( x, x ) x x 2 (1 τ) h(0, 0) ( +τ h(x η,x η ) + 12 x η ) x η 2 < (1 τ) h(0, 0) + τ η = h(0, 0) τ(h(0, 0) η). Therefore, using also (2.19) ε (h( x, x ) + 12 x ) x 2 ε h(0, 0) + τ(h(0, 0) η) which completes the proof. > ε h(0, 0) + τ(h(0, 0) 2η) = ε h(0, 0) + ( ) h(0, 0) h(0, 0) 2η = ε 2h(0, 0)η > 0. In Theorem the origin has a special role. In order to use this theorem with an arbitrary point, define [33], for h R X X and (z,z ) X X, h (z,z ) R X X, h (z,z )(x,x ) := h(x+z,x +z ) [ x,z + z,x + z,z ]. (2.21) Notice that h (z,z )(x,x ) x,x = h(x + z,x + z ) x + z,x + z. (2.22) The operation h h (z,z ) preserves many properties of h, as convexity, lower semicontinuity and can be seen as the action of the group (X X, +) on R X X, because ( h(z0,z 0 ) )(z 1,z 1 ) = h (z 0 +z 1,z 0 +z 1 ). The proof of Theorem will be heavily based on these nice properties of the map h h (z,z ). In the next proposition we prove X X that the class of l.s.c. convex functions h R such that h π and h π is invariant under the map h h (z,z ): Proposition (Marques Alves-Svaiter [29]) For any h it holds that: R X X 1. h is proper, convex and l.s.c. h (z,z ) is proper, convex and l.s.c., (z,z ) X X ; 15

25 2. ( h (z,z )) = (h ) (z,z), where the rightmost z is identified with its image under the canonical injection of X into X ; 3. h π, h π h (z,z ) π, ( h (z,z )) π, (z,z ) X X. Proof. Item 1 is trivial to check. For proving item 2, take (x,x ) X X. Then, using (2.21) we obtain ( ) h(z,z ) (x,x ) = sup y,x + y,x h (z,z )(y,y ) (y,y ) = h (x + z,x + z) [ x,z + z,x + z,z ] = (h ) (z,z)(x,x ). (2.23) It remains to prove item 3. For proving the if, note that h (0,0) = h. For proving the only if, use (2.22) to conclude that h (z,z ) π whenever h π. By the same reasoning, ( h ) (z,z) π whenever h π and so that using item 2 we end the proof of item 3. X X Corollary (Marques Alves-Svaiter [29]) Let h R be a convex and l.s.c. function. If h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, then for any (z,z ) X X and ε > 0 there exist ( x, x ) X X such that h( x, x ) < x, x + ε, x z 2 h(z,z ) z,z, x z 2 h(z,z ) z,z, where the two last inequalities are strict in the case z,z < h(z,z ). Proof. If h(z,z ) = z,z then ( x, x ) = (z,z ) satisfy the desired conditions. Assume that 0 < h(z,z ) z,z. (2.24) Using Proposition and Theorem we conclude that there exists (ỹ, ỹ ) X X such that h (z,z )(ỹ, ỹ ) ỹ ỹ 2 < ε, ỹ 2 < h (z,z )(0, 0), ỹ 2 < h (z,z )(0, 0). (2.25) 16

26 Using (2.22), we obtain h (z,z )(0, 0) = h(z,z ) z,z. Let x := ỹ + z, x := ỹ + z. Therefore, using (2.25) and (2.24), we have x z 2 < h(z,z ) z,z, x z 2 < h(z,z ) z,z. For finishing the proof of the corollary, use (2.22) and (2.25) to obtain h( x, x ) x, x = h (z,z )(ỹ, ỹ ) ỹ, ỹ h (z,z )(ỹ, ỹ ) ỹ ỹ 2 < ε. Now we come with the main result of this subsection. It is important by itself and will be used in the next sections, specially for proving Theorem X X Theorem (Marques Alves-Svaiter [29]) Let h R be convex, l.s.c. and h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X. If (x,x ) X X, ε > 0 and h(x,x ) < x,x + ε, then, for any λ > 0 there exists ( x λ, x λ ) X X such that h( x λ, x λ) = x λ, x λ, x λ x < λ, x λ x < ε λ. Proof. First, suppose that λ = ε. If h(x,x ) x,x = 0, then (x,x ) has the desired properties. So, suppose also that h(x,x ) x,x > 0. Let ε 0 > 0 and θ (0, 1) be such that 0 < h(x,x ) x,x ε0 < ε 0 < ε, 1 + θ < ε. (2.26) Define inductively a sequence {(x k,x k )} as follows: For k = 0, let (x 0,x 0) = (x,x ). (2.27) Given k and (x k,x k ), use Corollary to conclude that there exists some (x k+1,x k+1 ) such that h(x k+1,x k+1) x k+1,x k+1 < θ k+1 ε 0 (2.28) 17

27 and x k+1 x k h(x k,x k ) x k,x k, x k+1 x k h(x k,x k ) x k,x k. (2.29) Using (2.26) and (2.28) we conclude that for all k, which, combined with (2.29) yields x k+1 x k < ε 0 θk, k=0 0 h(x k,x k) x k,x k < θ k ε 0, (2.30) k=0 The second part of (2.26) gives x k+1 x k < ε, k=0 x k+1 x k < ε 0 θk. k=0 k=0 x k+1 x k < ε. (2.31) In particular, the sequences {x k } and {x k } are convergent. Let k=0 x := lim k x k, Then, using (2.31) we have x := lim k x k. Using (2.30) we have x x < ε, x x < ε. lim h(x k,x k) x k,x k = 0. k As h is l.s.c. and the duality product is continuous (in the strong topology of X X ), h( x, x ) x, x 0. Therefore, h( x, x ) x, x = 0, which ends the proof of the theorem for λ = ε. To prove the general case, use in X the norm ε x := λ x, and apply the previous case in this re-normed space Maximality of representable monotone operators Next we present one of the main results of this thesis. It generalizes Theorem for non-reflexive Banach spaces by replacing condition (1.12) by condition (1.13). 18

28 X X Theorem (Marques Alves-Svaiter [29]) Let h R be a convex and l.s.c. function. If then h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, T := {(x,x ) X X h (x,x ) = x,x } is maximal monotone and h, Λh F T. Proof. The duality product π : X X R is everywhere differentiable and π (x,x ) = (x,x). Suppose that h(x,x ) = x,x = π(x,x ). Then, by Lemma A.0.5 (x,x) h(x,x ), that is, h(x,x ) + h (x,x) = (x,x ), (x,x). (2.32) Take (x,x ), (y,y ) T. Then, as was explained above, Since h is monotone, (x,x) h(x,x ), (y,y) h(y,y ). (x,x ) (y,y ), (x,x) (y,y) 0, which gives x y,x y 0. Hence, T is monotone. Now, if h(x,x ) = x,x, then using (2.32) we have h (x,x) = x,x. Conversely, if h (x,x) = x,x, then by the same reasoning h (x,x ) = x,x. As h is proper, convex and l.s.c., h(x,x ) = h (x,x ). Thus, T = {(x,x ) X X h (x,x) = x,x }. (2.33) For proving maximal monotonicity of T, take (z,z ) X X and assume that x z,x z 0, (x,x ) T. (2.34) Using Theorem and Proposition we know that inf (x,x ) X X h (z,z )(x,x ) x x 2 = 0. Therefore, there exists a minimizing sequence {(x k,x k )} such that h (z,z )(x k,x k) x k x k 2 < 1 k2, k = 1, 2,... (2.35) 19

29 Note that the sequence {(x k,x k )} is bounded and h (z,z )(x k,x k) x k,x k h (z,z )(x k,x k) + x k x k h (z,z )(x k,x k) x k x k 2. Combining the two above inequalities we obtain h (z,z )(x k,x k) < x k,x k + 1 k 2. Now applying Theorem 2.2.4, we conclude that there for each k there exists some ( x k, x k ) such that h (z,z )( x k, x k) = x k, x k, x k x k < 1/k, x k x k < 1/k. Then, and from (2.34) (ȳ k,ȳ k) := ( x k + z, x k + z ) T, x k, x k = ȳ k z,ȳ k z 0. The duality product is uniformly continuous on bounded sets. Since {(x k,x k )} is bounded and lim k x k x k = lim k x k x k = 0 we conclude that lim inf k x k,x k 0. Using (2.35) and the fact that h majorizes the duality product, we have 0 x k,x k x k x k 2 h (z,z )(x k,x k)+ 1 2 x k x k 2 < 1 k 2. Hence, x k,x k < 1/k2 and lim sup k x k,x k 0, which implies lim k x k,x k = 0. Combining this result with the above inequalities we conclude that lim (x k,x k) = 0. k Therefore, lim k ( x k, x k ) = 0 and {(ȳ k,ȳk )} converges to (z,z ). As h(ȳ k,ȳk ) = ȳ k,ȳk and h is lower semicontinuous, h(z,z ) z,z. which readily implies h(z,z ) = z,z. Therefore (z,z ) T and so that T is maximal monotone and h F T. Finally, using (2.33) we conclude that Λh F T. 20

30 The duality product is continuous in X X. Therefore, if a convex function majorizes the duality product then the convex closure of this function also majorizes it and has the same conjugate. This fact can be used to remove the assumption of lower semicontinuity of h in Theorem 2.2.5: X X Theorem (Marques Alves-Svaiter [29]) Let h R be a convex function. If then h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, S := {(x,x ) X X h (x,x) = x,x } is maximal monotone and cl h, Λh F S. (2.36) Proof. Define h := cl h. Then, h is proper, convex, l.s.c. and ( h) = h. Since the duality product is continuous, h also satisfies (2.36). Thus, applying Theorem to h, we have that S is maximal monotone and clh, Λh F S Brøndsted-Rockafellar type theorems for representable and of type (NI) operators The class of maximal monotone operators of type (NI) was introduced by S. Simons for generalizing some results in reflexive Banach spaces to non-reflexive spaces: Definition ( [39]) A maximal monotone operator T : X X is of type (NI) if inf y x,y x 0, (x,x ) X X. (y,y ) T As pointed out before, in this subsection we aim to prove that maximal monotone operators representable by convex function satisfying (1.13) and operators of type (NI) satisfy the strict Brøndsted- Rockafellar property. X X Theorem (Marques Alves-Svaiter [29]) Let h R be a convex and l.s.c. function. If h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, then the maximal monotone operator (see Theorem 2.2.5) T := {(x,x ) X X h (x,x ) = x,x } 21

31 satisfies the strict Brøndsted-Rockafellar property: If η > ε and x T ε (x), that is, x y,x y ε, (y,y ) T, then, for any λ > 0 there exists ( x λ, x λ ) X X such that x λ T( x λ ), x x λ < λ, x x λ < η λ. Proof. Assume that η > ε > 0 and x y,x y ε, (y,y ) T. The Fitzpatrick function of T is Therefore ϕ T (x,x ) = sup x,y + y,x y,y (y,y ) T = sup (y,y ) T x y,x y + x,x. ϕ T (x,x ) x,x + ε < x,x + η. Now recall that, as T is maximal monotone, ϕ T is the smallest element of the family F T. In particular h ϕ T. Hence, ϕ T h, which implies that ϕ T satisfies the hypothesis of Theorem Thus, there exists ( x λ, x λ ) such that ϕ T ( x λ, x λ) = x λ, x λ, x x λ < λ, x x λ < η λ. The first equality above says that ( x λ, x λ ) T, which ends the proof of the theorem. Theorem (Marques Alves-Svaiter [29]) Let h convex function. If R X X be a h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X, then the maximal monotone operator (see Theorem 2.2.6) S := {(x,x ) X X h (x,x) = x,x } 22

32 satisfies the strict Brøndsted-Rockafellar property: If η > ε and x S ε (x), that is, x y,x y ε, (y,y ) S, then, for any λ > 0 there exists ( x λ, x λ ) X X such that x λ S( x λ ), x x λ < λ, x x λ < η λ. In Theorem we shall prove that the class of maximal monotone operators of type (NI) satisfies the strict Brøndsted-Rockafellar property. The starting point of the proof is a characterization of the class of operators of type (NI) given by the S-function: Proposition (Marques Alves-Svaiter [31]) A maximal monotone operator T : X X is of type (NI) if, and only if, (S T ) (x,x ) x,x, (x,x ) X X. Proof. Recall that S T = clconv(π + δ T ). The proof follows directly from the identity below: (S T ) (x,x ) = (π + δ T ) (x,x ) = sup y,x + y,x y,y (y,y ) T = inf x y,x y + x,x (y,y ) T Theorem (Marques Alves-Svaiter [31]) Let T : X X be a maximal monotone operator of type (NI). Then, T satisfies the strict Brøndsted-Rockafellar property: If η > ε and x T ε (x), that is, x y,x y ε, (y,y ) T, then, for any λ > 0 there exists ( x λ, x λ ) X X such that x λ T( x λ ), x x λ < λ, x x λ < η λ. Proof. Recall that S T F T and so T = {(x,x ) X X S T (x,x ) = x,x }. Using the fact that S T F T and Proposition 2.2.9, we have that S T (x,x ) x,x, (x,x ) X X, (S T ) (x,x ) x,x, (x,x ) X X, Thus, the result follows from Theorem

33 2.3 A new characterization and a sum theorem for operators of type (NI) Proposition says that if a maximal monotone operator T : X X is of type (NI) then there exits h F T, namely h = S T, such that h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X. (2.37) In Theorem we used the fact that if h satisfies condition (2.37), then it also satisfies the following variational condition: inf (x,x ) h (x 0,x 0 ) (x,x ) x x 2 = 0, (x 0,x 0). (2.38) In Theorem we will show that conditions (2.37) and (2.38) are equivalent and that if some h F T satisfies condition (2.37), then all function in the Fitzpatrick family of T satisfies condition (2.37). Remember that if h F T, then h π holds by definition of F T. In particular, (2.38) provides a sort of variational characterization of the class of maximal monotone operators of type (NI). Theorem (Marques Alves-Svaiter [32]) Let T : X X be maximal monotone. The following conditions are equivalent: 1. T is of type (NI), 2. there exists h F T such that 3. for all h F T, h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X. h(x,x ) x,x, (x,x ) X X, h (x,x ) x,x, (x,x ) X X. 4. there exists h F T such that inf h (x0 (x,x,x 0 ) ) (x,x )+ 1 2 x x 2 = 0, (x 0,x 0) X X, 5. for all h F T, inf h (x,x (x 0,x 0 ) ) (x,x )+ 1 2 x x 2 = 0, (x 0,x 0) X X. 24

34 Proof. First let us prove that item 2 and item 4 are equivalent. Using Theorem we conclude that item 2 implies item 4. For proving that item 4 implies item 2, first note that, for any (z,z ) X X, h (z,z )(0, 0) inf h (z,z )(x,x ) + 1 (x,x ) 2 x x 2. Therefore, using item 4 we obtain h(z,z ) z,z = h (z,z )(0, 0) 0. Since (z,z ) is an arbitrary element of X X we conclude that h π. For proving that h π, take some (y,y ) X X. First, use Fenchel-Young inequality to conclude that for any (x,x ), (z,z ) X X, h (z,z )(x,x ) x,y z + x,y z ( h (z,z )) (y z,y z). As ( h (z,z )) = (h ) (z,z), ( ) h(z,z ) (y z,y z) = h (y,y ) y,y + y z,y z. Combining the two above equations we obtain h (z,z )(x,x ) x,y z + x,y z y z,y z + y,y h (y,y ). Adding (1/2) x 2 +(1/2) x 2 in both sides of the above inequality we have h (z,z )(x,x ) x x 2 x,y z + x,y z x x 2 Note that y z,y z + y,y h (y,y ). x,y z x y z 2, x,y z x y z 2. Therefore, for any (x,x ), (z,z ) X X, h (z,z )(x,x ) x x y z y z 2 y z,y z + y,y h (y,y ). Using now the assumption we conclude that the infimum, for (x,x ) X X, at the left hand side of the above inequality is 0. Therefore, 25

35 taking the infimum on (x,x ) X X at the left hand side of the above inequality and rearranging the resulting inequality we have h (y,y ) y,y 1 2 y z y z 2 y z,y z. Note that sup z X y z,y z 1 2 y z 2 = 1 2 y z 2. Hence, taking the sup in z X at the right hand side of the previous inequality, we obtain h (y,y ) y,y 0 and item 4 holds. Now, using that item 2 and item 4 are equivalent it is trivial to verify that item 3 and item 5 are equivalent. The second step is to prove that item 4 and item 5 are equivalent. So, assume that item 4 holds, that is, for some h F T, inf h (x,x ) X X (x 0,x 0 ) (x,x )+ 1 2 x x 2 = 0, (x 0,x 0) X X. Take g F T, and (x 0,x 0) X X. First observe that, for any (x,x ) X X, g (x0,x 0 ) (x,x ) x,x and g (x0,x 0 ) (x,x ) x x 2 x,x x x 2 0. Therefore, inf (x,x ) X X g (x 0,x 0 ) (x,x ) x x 2 0. (2.39) As the square of the norm is coercive, there exist M > 0 such that { (x,x ) X X h (x0,x 0 ) (x,x ) x } 2 x 2 < 1 B X X (0,M), where B X X (0,M) = { } (x,x ) X X x 2 + x 2 < M. For any ε > 0, there exists ( x, x ) such that min { 1,ε 2} > h (x0,x 0 ) ( x, x ) x x 2. 26

36 Therefore ε 2 > h (x0,x 0 ) ( x, x ) x x 2 h (x0,x 0 ) ( x, x ) x, x 0, M 2 x 2 + x 2. In particular, ε 2 > h (x0,x 0 ) ( x, x ) x, x. (2.40) Now using Theorem we conclude that there exists ( x, x ) such that h (x0,x 0 ) ( x, x ) = x, x, x x < ε, x x < ε. (2.41) Therefore, h( x + x 0, x + x 0) x + x 0, x + x 0 = h (x0,x 0 ) ( x, x ) x, x = 0, and ( x + x 0, x + x 0) T. As g F T, and g( x + x 0, x + x 0) = x + x 0, x + x 0, g (x0,x 0 ) ( x, x ) = x, x. (2.42) Using the first line of (2.40) we have [ 1 ε 2 > h (x0,x 0 ) ( x, x )+ 2 x ] 2 x 2 + x, x x, x 1 2 x x 2 + x, x. Therefore, Direct use of (2.41) gives ε 2 > 1 2 x x 2 + x, x. (2.43) x, x = x, x + x x, x + x, x x + x x, x x x, x + x x x + x x x + x x x x and x, x + ε[ x + x ] + ε 2 x 2 + x 2 ( x + x x ) 2 + ( x + x x ) 2 x 2 + x 2 + 2ε[ x + x ] + 2ε 2 Combining the two above equations with (2.42) we obtain g (x0,x 0 ) ( x, x )+ 1 2 x x 2 x, x x x 2 +2ε[ x + x ]+2ε 2 27

37 Using now (2.43) and the second line of (2.40) we conclude that g (x0,x 0 ) ( x, x ) x x 2 2ε M 2 + 3ε 2. As ε is an arbitrary strictly positive number, using also (2.39) we conclude that inf (x,x ) X X g (x 0,x 0 ) (x,x ) x x 2 = 0. Altogether, we conclude that if item 4 holds then item 5 holds. The converse (item 5 implies item 4) is trivial to verify. Hence item 4 and item 5 are equivalent. As item 2 is equivalent to item 4 and item 3 is equivalent to 5, we conclude that items 2,3,4 and 5 are equivalent. Now we will prove that item 1 is equivalent to item 3 and conclude the proof of the theorem. First suppose that item 3 holds. Since S T F T (S T ) π. As has already been observed, for any proper function h it holds that (cl convh) = h. Therefore that is, (S T ) = (π + δ T ) π, sup y,x + y,x y,y x,x, (x,x ) X X (y,y ) T (2.44) After some algebraic manipulations we conclude that (2.44) is equivalent to inf x y,x y 0, (x,x ) X X, (y,y ) T that is, T is type (NI) and so item 1 holds. If item 1 holds, by the same reasoning we conclude that (2.44) holds and therefore (S T ) π. As S T F T, we conclude that item 2 holds. As has been proved previously item 2 item 3. In the next lemma we give a sufficient condition for proving that the sum of maximal monotone operators of type (NI) is of type (NI). Lemma (Marques Alves-Svaiter [30]) Let T 1,T 2 : X X be maximal monotone and of type (NI). Take h 1 F T1, h 2 F T2 28

38 and define h R X X, h(x,x ) := (h 1 (x, ) h 2 (x, )) (x ) = inf y X h 1(x,y ) + h 2 (x,x y ), If Pr X dom(h i ) := {x X x, h i (x,x ) < }, i = 1, 2. λ [Pr X dom(h 1 ) Pr X dom(h 2 )] (2.45) λ>0 is a closed subspace then h π,h π, Λh π, (Λh) π, T 1 + T 2 = {(x,x ) Λh(x,x ) = x,x } = {(x,x ) h(x,x ) = x,x } and T 1 + T 2 is maximal monotone of type (NI) and Λh, cl h F T1 +T 2. Proof. Since h 1 F T1 and h 2 F T2, h 1 π and h 2 π. So h 1 (x,y ) + h 2 (x,x y ) x,y + x,x y = x,x. Taking the inf in y at the left-hand side of the above inequality we conclude that h π. Let (x,x ) X X. Using the definition of h we have h (x,x ) = sup (z,z ) X X z,x + z,x h(z,z ) (2.46) = sup (z,z,y ) X X X z,x + z,x h 1 (z,y ) h 2 (z,z y ) (2.47) = sup (z,y,w ) X X X z,x + y,x + w,x h 1 (z,y ) h 2 (z,w ) (2.48) where we used the substitution z = w + y in the last term. So, defining H 1,H 2 : X X X R H 1 (x,y,z ) = h 1 (x,y ), H 2 (x,y,z ) = h 2 (x,z ). (2.49) we have h (x,x ) = (H 1 + H 2 ) (x,x,x ). 29

39 Using (2.45), Theorem A.0.4 and (2.49) we conclude that the conjugate of the sum at the right hand side of the above equation is the exact inf-convolution of the conjugates. Therefore, h (x,x ) = min (u,y,z ) H 1(u,y,z )+H2(x u,x y,x z ). Direct use of definition (2.49) yields H 1(u,y,z ) = h 1(u,y )+δ 0 (z ), (u,y,z ) X X X, (2.50) H 2(u,y,z ) = h 2(u,z )+δ 0 (y ), (u,y,z ) X X X. (2.51) Hence, h (x,x ) = min u X h 1(u,x ) + h 2(x u,x ). (2.52) Therefore, using that h 1 π, h 2 π, (2.52) and the same reasoning used to show that h π we have h π. Up to now, we proved that h π and h π ( and Λh π). So, using Theorem we conclude that S : X X, defined as S = {(x,x ) X X Λh(x,x ) = x,x }, is maximal monotone. Since Λh is convex and lower semicontinuous, Λh F S. We will prove that T 1 + T 2 = S. Take (x,x ) S, that is, Λh(x,x ) = x,x. Using (2.52) we conclude that there exists u X such that We know that h 1(u,x) + h 2(x u,x) = x,x. h 1(u,x) x,u, h 2(x u,x) x,x u. Combining these inequalities with the previous equation we conclude that these inequalities hold as equalities, and so u T 1 (x), x u T 2 (x), x (T 1 + T 2 )(x). h 1 (x,u ) = x,u, h 2 (x,x u ) = x,x u, h(x,x ) x,x. We proved that S T 1 + T 2. Since T 1 + T 2 is monotone and S is maximal monotone, we have T 1 + T 2 = S (and Λh F T1 +T 2 ). Note 30

40 also that h(x,x ) x,x for any (x,x ) T 1 +T 2 = S. As h π, we have equality in T 1 + T 2. Therefore, T 1 +T 2 {(x,x ) h(x,x ) = x,x } {(x,x ) cl h(x,x ) x,x }. Since h π and the duality product π is continuous in X X, we also have clh π. Hence, using the above inclusion we conclude that clh coincides with π in T 1 + T 2. Therefore, clh F T1 +T 2 and the rightmost set in the above inclusions is T 1 + T 2. Hence T 1 + T 2 = {(x,x ) h(x,x ) = x,x }. Conjugation is invariant under the (lower semicontinuous) closure operation. Therefore, (cl h) = h π and so T 1 +T 2 is of type (NI). We proved already that Λh F T1 +T 2. Using item 3 of Theorem we conclude that (Λh) π. 31

41 Chapter 3 On the relation between surjectivity of perturbations and operators of type (NI) In this chapter we are concerned with surjectivity of perturbation of maximal monotone operators in non-reflexive Banach spaces. The results presented here are collected from [30]. In a reflexive Banach space the following result due to Rockafellar gives a necessary and sufficient condition for maximal monotonicity in terms of the surjectivity of perturbations by the duality mapping J: Theorem ([38, Proposition 1]). Let X be a reflexive Banach space and T : X X be a monotone operator. Then T is maximal monotone if and only if R(T( + z 0 ) + J) = X, z 0 X. Here, by (x,x ) T( + z 0 ) we means (x + z 0,x ) T. Recall that the duality mapping is the point to set operator J : X X defined by J(x) = 1 2 x 2. The point is that J is surjectivity if and only if X is reflexive. In order to overcome this difficult, J.-P. Gossez introduced [21] an enlarged version of the duality mapping, J ε : X X defined by J ε (x) = ε 1 2 x 2, and obtained similar resuts of Theorem for a special class of maximal monotone operators he introduced in non-reflexive Banach 32

42 spaces, the operators of type (D). Notice that for any ε > 0, J ε is always surjectivity. Recall that a maximal monotone operator T : X X is of type (NI) if inf y x,y x 0, (x,x ) X X. (y,y ) T The class of maximal monotone operators of type (NI) encompasses the Gossez type (D) operators and was introduced by S. Simons [39] to generalize some results concerning maximal monotonicity in reflexive Banach spaces for non-reflexive Banach spaces. The general framework of convex representations of maximal monotone operators developted in the previous chapters allows us to characterize the operators of type (NI) by the existence of a Fitzpatrick function in the Fitzpatrick family such that the conjugate majorizes the duality product (see Theorem 2.3.1). In the next sections, we will use this results to obtain surjectivity results for perturbations of maximal monotone operators of type (NI). 3.1 Preliminary results We begin with two elementary technical results which will be useful. Proposition (Marques Alves-Svaiter [30]) The following statements holds: 1. For any ε 0, if y J ε (x), then x y 2ε. 2. Let T : X X be a monotone operator and ε,m > 0. Then, is bounded. (T + J ε ) 1 (B X [0,M]) Proof. For proving item 1, let ε 0 and y J ε (x). The desired result follows from the following inequalities: 1 2 ( x y ) x y 2 x,y ε. For proving item 2, take (z,z ) T. If x (T + J ε ) 1 (B[0,M]) then there exists x,y such that x T(x), y J ε (x), x + y M. 33

43 Therefore, using Fenchel Young inequality (A.2), the monotonicity of T and the definition of J ε we obtain 1 2 x z x + y z 2 x z,x + y z Note also that x z,y [ 1 2 x ] 2 y 2 ε z y. x z 2 x 2 +2 x z + z 2, x + y z 2 (M+ z ) 2. Combining the above equations we obtain 1 2 z (M + z ) y 2 x z z y ε. As y J ε (x), by item 1, we have x y + 2ε. Therefore 1 2 z (M + z ) y 2 2 y z z 2ε ε. Hence, y is bounded. In fact, [ y 2 z + 4 z z ] 2ε + ε + z 2 + (M + z ) 2. As we already observed, x y + 2ε and so, x is also bounded. Now we will prove that under monotonicity, dense range of some perturbation of a monotone operator is equivalent to surjectivity of that perturbation. Lemma (Marques Alves-Svaiter [30]) Let T : X X be monotone and µ > 0. Then the conditions below are equivalent 1. R(T( + z 0 ) + µj ε ) = X, for any ε > 0 and z 0 X, 2. R(T( + z 0 ) + µj ε ) = X, for any ε > 0 and z 0 X. Proof. It suffices to prove the lemma for µ = 1 and then, for the general case, consider T = µ 1 T. Now note that for any z 0 X and z 0 X, T {(z 0,z 0)} is also monotone. Therefore, it suffices to prove that 0 R(T + J ε ), for any ε > 0 if and only if 0 R(T +J ε ), for any ε > 0. The if is easy to check. To prove the only if, suppose that 0 R(T + J ε ), ε > 0. 34

44 First use item 2 of Proposition with M = 1/2 to conclude that there exists ρ > 0 such that (T + J 1/2 ) 1 (B X [0, 1/2]) B X [0,ρ]. By assumption, for any 0 < η < 1 2 there exists x η X, x η,y η X such that x η T(x η ), y η J η (x η ) and x η + y η < η < 1 2. (3.1) As J η (x η ) J 1/2 (x η ), x η (T + J 1/2 ) 1 (x η + y η) and so, x η ρ, y η ρ + 1. where the second inequality follows from the first one and item 1 of Proposition Therefore 1 2 x η 2 1 ( x 2 η + yη + yη ) η2 + η(ρ + 1) y η 2, x η,x η = x η,x η + y η x η,y η ρη x η,y η. Combining the above inequalities we obtain 1 2 x η x η 2 + x η,x η 1 2 x η y η 2 x η,y η + η(2ρ + 1) η2. The inclusion y η J η (x η ), means that, 1 2 x η y η 2 x η,y η η. (3.2) Hence, using the two above inequalities we conclude that 1 2 x η x η 2 + x η,x η 2η(ρ + 1) η2. For finishing the prove, take an arbitrary ε > 0. Choosing 0 < η < 1/2 such that, we have 2η(ρ + 1) η2 < ε, 1 2 x η x η 2 + x η,x η < ε, x η T(x η ). According tho the above inequality, x η J ε (x η ). Hence 0 (T + J ε )(x η ). 35